Search
You can find the results of your search below.
Matching pagenames:
- FSc Part 1 Mathematics Notes/Solutions
- Solution and Area of Oblique Triangle
- FSc Part 1 Mathematics Notes/Solutions
- FSc/ICS Part 2 Solutions
- Solutions: Math 11 KPK
- Vector & Tensor Analysis by Dr Nawazish Ali (Solutions)
- Unit 1: Complex Numbers (Solutions)
- Unit 10: Trigonometric Identities of Sum and Difference of Angles (Solutions)
- Chapter 01: Number System
- Chapter 04: Quadratic Equations
- Chapter 09: Fundamentals of Trigonometry
- Unit 01: Functions and Limits
- Unit 03: Integration
- Unit 01: Complex Numbers (Solutions)
- Unit 02: Matrices and Determinants (Solutions)
- Unit 03: Vectors (Solutions)
- Unit 04: Sequence and Series (Solutions)
- Unit 05: Miscellaneous Series (Solutions)
- Unit 06: Permutation, Combination and Probability (Solutions)
- Unit 07: Mathmatical Induction and Binomial Theorem (Solutions)
- Unit 10: Trigonometric Identities of Sum and Difference of Angles (Solutions)
- Mathematical Method by Khalid Latif Mir (Solutions)
- Unit 02: Differentiation: Mathematics FSc part 2
Fulltext results:
- Question 7, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- \sin }^{4}}\theta =\dfrac{1}{\sec 2\theta }$. ====Solution==== \begin{align}L.H.S&={{\cos }^{4}}\theta -{{\s... }\dfrac{\theta }{2}=\dfrac{2}{\sin \theta }$. ====Solution==== \begin{align}L.H.S&=\tan \dfrac{\theta }{2}+c... eta }{1-\cos 2\theta }={{\cot }^{2}}\theta $. ====Solution==== \begin{align}L.H.S&=\dfrac{1+\cos 2\theta }{1... y $\cos ec2\theta -\cot 2\theta =tan\theta $. ====Solution==== \begin{align}L.H.S.&=\cos ec2\theta -\cot 2\t
- Question 7, Exercise 10.2 @math-11-kpk:sol:unit10
- \sin }^{4}}\theta =\dfrac{1}{\sec 2\theta }$. ====Solution==== \begin{align}L.H.S&={{\cos }^{4}}\theta -{{\s... }\dfrac{\theta }{2}=\dfrac{2}{\sin \theta }$. ====Solution==== \begin{align}L.H.S&=\tan \dfrac{\theta }{2}+c... eta }{1-\cos 2\theta }={{\cot }^{2}}\theta $. ====Solution==== \begin{align}L.H.S&=\dfrac{1+\cos 2\theta }{1... y $\cos ec2\theta -\cot 2\theta =tan\theta $. ====Solution==== \begin{align}L.H.S.&=\cos ec2\theta -\cot 2\t
- Question 7 Exercise 6.4 @math-11-kpk:sol:unit06
- obability of getting doublet of even numbers. ====Solution==== The sample space rolling a pair of dice is \b... probability of getting a sum less than $6$. ====Solution==== The sample space rolling a pair of dice is \b... e probability of getting a sum more than $7.$ ====Solution==== The sample space rolling a pair of dice is \b... bability of getting a sum greater than $10.$ ====Solution==== The sample space rolling a pair of dice is \b
- Operation Research: Handwritten Notes @notes
- jective function 2 * Decision variables 2 * Solution space 2 * Optimum solution 2 * Standard form 2 * Conversion of maximization to minimization 3 * Linear programming formulation and graphical solution 4 <grid> <col sm="6"> * Simplex method (or the general solution technique) 22 * Algorithm 22 * Optimality c
- Question 1 and 2 Exercise 4.1 @math-11-kpk:sol:unit04
- nd infinite sequences\\ $2,4,6,8, \ldots ,50$ ====Solution==== It is finite sequence whose last term is $50... and infinite sequences. $1,0,1,0,1, \ldots$. ====Solution==== It is infinite sequence, the last term may be... nfinite sequences: $...,-4,0,4,8, \ldots, 60$ ====Solution==== This is infinite sequence. =====Question 1(i... }{9},-\dfrac{1}{27}, \ldots,-\dfrac{1}{2187}$ ====Solution==== This finite sequence. =====Question 2(i)====
- Question 1 and 2 Exercise 6.1 @math-11-kpk:sol:unit06
- aluate the $\dfrac{10 !}{3 ! .3 ! \cdot 4 !}$ ====Solution==== \begin{align}\dfrac{10 !}{3 ! \cdot 3 ! \cdot... ===== Evaluate the $\dfrac{3 !+4 !}{5 !-4 !}$ ====Solution==== \begin{align}\dfrac{3 !+4 !}{5 !-4 !}&=\dfrac... ===== Evaluate the $\dfrac{(n-1) !}{(n+1) !}$ ====Solution==== \begin{align}\dfrac{(n-1) !}{(n+1) !} &= \df... iv)===== Evaluate the $\dfrac{10 !}{(5 !)^2}$ ====Solution==== \begin{align}\dfrac{10 !}{(5 !)^2}&=\dfrac{10
- Question 7, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- eal and imaginary parts $\dfrac{2+3i}{5-2i}$. ====Solution==== \begin{align}&\dfrac{2+3i}{5-2i} \\ =&\dfrac... $\dfrac{{{\left( 1+2i \right)}^{2}}}{1-3i}$. ====Solution==== \begin{align}&\dfrac{(1+2i)^2}{1-3i}\\ =&\df... ts $\dfrac{1-i}{{{\left( 1+i \right)}^{2}}}$. ====Solution==== \begin{align}&\dfrac{1-i}{{{\left( 1+i \righ... ginary parts ${{\left( 2a-bi \right)}^{-2}}$. ====Solution==== \begin{align}&{{\left( 2a-bi \right)}^{-2}}\
- Question 1, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- }}+\cos {{37}^{\circ }}\sin {{22}^{\circ }}$ ==== Solution ==== As \begin{align} \sin (\alpha +\beta )=\si... }}+\sin {{83}^{\circ }}\sin {{53}^{\circ }}$ ====Solution==== As \begin{align}\cos (\alpha -\beta )&=\cos... }}-\sin {{19}^{\circ }}\sin {{5}^{\circ }}$ ==== Solution ==== As \begin{align}\cos (\alpha +\beta )&=\cos... }}-\cos {{40}^{\circ }}\sin {{15}^{\circ }}$ ==== Solution ==== As \begin{align}\sin (\alpha -\beta )=\sin
- Question 2, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- === Evaluate exactly: $\sin \dfrac{\pi }{12}$ ===Solution=== We rewrite $\dfrac{\pi }{12}$ as $\dfrac{\pi }... ii)=== Evaluate exactly:$\tan {{75}^{\circ }}$ ==Solution== We rewrite ${{75}^{\circ }}$ as ${{45}^{\circ }... i)=== Evaluate exactly:$\tan {{105}^{\circ }}$ ==Solution== We rewrite ${{105}^{{}^\circ }}$ as ${{60}^{{}^... v)=== Evaluate exactly:$\tan \dfrac{5\pi }{12}$ ==Solution== We rewrite $\dfrac{5\pi }{12}$ as $\dfrac{\left
- Question 6, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- s to evaluate exactly $\cos {{15}^{\circ }}$. ====Solution==== Because ${{15}^{\circ }}=\dfrac{{{30}^{\circ ... to evaluate exactly $\tan {{67.5}^{\circ }}$. ====Solution==== Because ${{67.5}^{\circ }}=\dfrac{{{135}^{\ci... to evaluate exactly $sin{{112.5}^{\circ }}$. ====Solution==== Because ${{112.5}^{\circ }}=\dfrac{{{225}^{\c... s to evaluate exactly $\cos \dfrac{\pi }{8}$. ====Solution==== Because $\dfrac{\pi }{8}=\dfrac{\dfrac{\pi }{
- Question 7, Exercise 1.2 @math-11-kpk:sol:unit01
- eal and imaginary parts $\dfrac{2+3i}{5-2i}$. ====Solution==== \begin{align}&\dfrac{2+3i}{5-2i} \\ =&\dfrac... $\dfrac{{{\left( 1+2i \right)}^{2}}}{1-3i}$. ====Solution==== \begin{align}&\dfrac{(1+2i)^2}{1-3i}\\ =&\df... ts $\dfrac{1-i}{{{\left( 1+i \right)}^{2}}}$. ====Solution==== \begin{align}&\dfrac{1-i}{{{\left( 1+i \righ... ginary parts ${{\left( 2a-bi \right)}^{-2}}$. ====Solution==== \begin{align}&{{\left( 2a-bi \right)}^{-2}}\
- Question 2, Exercise 2.2 @math-11-kpk:sol:unit02
- & 1 & 0 \\-1 & 2 & 0 \end{matrix}\right|=0$. ====Solution==== Given $$\left| \begin{matrix} 1 & 2 & 0 \... & -12 \\2 & -1 & 3 \end{matrix} \right|=0$. ====Solution==== Given $$\left| \begin{matrix} 1 & 2 & 3 \... & -1 & 1 \\-2 & 1 & 4 \end{matrix} \right|$. ====Solution==== Given $$\left| \begin{matrix} 1 & 3 & -2 ... 1 & 1 & 1 \\2 & 4 & 2 \end{matrix} \right|$. ====Solution==== Given $$\left| \begin{matrix}3 & 2 & 0 \\1 &
- Question 1, Exercise 3.2 @math-11-kpk:sol:unit03
- i}+3\hat{j}$, then find $\vec{a}+2\vec{b}$. ====Solution==== \begin{align}\vec{a}+2\vec{b}&=3\hat{i}-5\hat... }+3\hat{j}$, then find $3\vec{a}-2\vec{b}$. ====Solution==== \begin{align}3\vec{a}-2\vec{b}&=3(3\hat{i}-5\... +3\hat{j}$, then find $2(\vec{a}-\vec{b})$. ====Solution==== First we have, \begin{align}\vec{a}-\vec{b}&=... }+3\hat{j}$, then find $|\vec{a}+\vec{b}|$. ====Solution==== We have, \begin{align}\vec{a}+\vec{b}&=3\hat{
- Question 1 Exercise 4.5 @math-11-kpk:sol:unit04
- i)===== Compute the sum $3+6+12+\ldots+3.2^9$ ====Solution==== In the given geometric series: $a_1=3, \quad ... Compute the sum $8+4+2+\ldots+\dfrac{1}{16}$ ====Solution==== In the give geometric series $$a_1=8, \quad r... = Compute the sum $2^4+2^5+2^6+\ldots+2^{10}$ ====Solution==== In the give geometric series.\\ $$a_1=2^4, \q... sum $\dfrac{8}{5},-1, \dfrac{5}{8}, \ldots$, ====Solution==== Here $$a_1=\dfrac{8}{5}$$ \begin{align}r&=\d
- Question 1 and 2 Exercise 6.2 @math-11-kpk:sol:unit06
- n. =====Question 1(i)===== Evaluate $^6 P_6$ ====Solution==== \begin{align}^6 P_6&=\dfrac{6 !}{(6-6) !}\\ &... =====Question 1(ii)===== Evaluate $^{20} P_2$ ====Solution==== \begin{align}^{20} P_2&=\dfrac{20 !}{(20-2) !... ====Question 1(iii)===== Evaluate $^{16} P_3$ ====Solution==== \begin{align}^{16} P_3&=\dfrac{16 !}{(16-3) !... 2(i)===== Solve $^n P_5=56(^n P_3)$ for $n.$ ====Solution==== We are given: \begin{align}^n P_5&=56(^n P_3)